Implicit Differentiation of an equation.

Guys, I'm having a hard time on understanding the Implicit Differentiation of an equation. Now I'm asking some help to all of you. I will give an equation and KINDLY tell it to me how you got the answer :(..Thanks

Assuming that the equation determines a differentiable function f such that y=f(x), find y'

1. 2x^3+x^2y+y^3=1

2.√x+√y=100

3. xy=tan y

Thanks for those who will answer and I would be glad if you can explain to me how you got the answer.. :)

Re: Implicit Differentiation of an equation.

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Originally Posted by

**coen000** Guys, I'm having a hard time on understanding the Implicit Differentiation of an equation. Now I'm asking some help to all of you. I will give an equation and KINDLY tell it to me how you got the answer :(..Thanks

Assuming that the equation determines a differentiable function f such that y=f(x), find y'

1. 2x^3+x^2y+y^3=1

The chain rule: if f(y) is a function of y and y is a function of x, then $\displaystyle \frac{df}{dx}= \frac{df}{dy}\frac{dy}{dx}= \frac{df}dy}y'$. Anytime you differentiate an expression in y, with respect to x, differentiate with respect to y and multiply by y'.

By the chain rule, the deirivative of $\displaystyle x^2y$, with respect to x, is $\displaystyle 2xy+ x^2y'$. The derivative of $\displaystyle y^3$, with respect to x, is $\displaystyle 3y^2y'$. Of course, the derivative of $\displaystyle 2x^3$, with respect to x, is $\displaystyle 6x^2$. The derivative of the constant, 1, is, of course, 0. Put those together and solve for y'.

$\displaystyle \sqrt{x}+ \sqrt{y}= x^{1/2}+ y^{1/2}= 1$

The derivative of $\displaystyle x^{1/2}$, with respect to x, is $\displaystyle \frac{1}{2}x^{-1/2}$ and the derivative of $\displaystyle y^{1/2}$, with respect to x, is [tex]\frac{1}{2}y^{-1/2}y'[tex]

Solve $\displaystyle \frac{1}{2}x^{-1/2}+ \frac{1}{2}y^{-1/2}y'= 0$ for y'.

The derivative of xy, with respect to x, is y+ xy'. The derivative of tan(y), with respect to x, is $\displaystyle sec^2(y)y'$.

Solve $\displaystyle y+ xy'= sec^2(x)y'$ for y'.

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Thanks for those who will answer and I would be glad if you can explain to me how you got the answer.. :)